find the lim as x approaches 0 for sin(sin(sin(kx)))/x
I used l'Hopital's rule and got this
lim as x approaches 0 for cos(sin(sin(kx)cos(sin(kx))kcos(kx)
the answer is 0.
The reason this didn't work is that you are missing a term in the above. Let $\displaystyle f(x)=\frac{sin [ sin ( sin[kx] ) ]}{x} $ thenOriginally Posted by Nichelle14
$\displaystyle f'(x) = \frac{cos [ sin ( sin [kx] ) ] cos ( sin[kx] ) cos[kx] k}{x}$ - $\displaystyle \frac{sin [ sin ( sin[kx] ) ]}{x^2}$
This expression gives a mess if you try to take the limit as x goes to 0. (You wind up doing an infinite number of L'Hopital rules.)
My suggestion is to use a nested series of approximations for small values of x in the sine function:
$\displaystyle \lim_{x \to 0}sin(x) \to x - (1/6)x^3$
and keep terms to third order in x.
So the numerator of your limit becomes:
$\displaystyle \lim_{x \to 0}sin [ sin ( sin[kx] ) ] \to \lim_{x \to 0} sin [ sin ( kx - (1/6)k^3x^3 ) ] $ $\displaystyle \to \lim_{x \to 0}sin [ \Big(kx - (1/6)k^3x^3 \Big) - (1/6) \Big(kx - (1/6)k^3x^3 \Big)^3 ] $
$\displaystyle \to \lim_{x \to 0}sin [ kx - (1/6)k^3x^3 - (1/6)k^3x^3 ] $ (to third order in x)
$\displaystyle \to \lim_{x \to 0}sin [ kx - (2/6)k^3x^3 ] $
$\displaystyle \to \lim_{x \to 0}[ \Big( kx - (2/6)k^3x^3 \Big) - (1/6) \Big( kx - (2/6)k^3x^3 \Big)^3] $ $\displaystyle \to \lim_{x \to 0}[kx - (2/6)k^3x^3 - (1/6)k^3x^3] $ $\displaystyle \to \lim_{x \to 0}[kx - (1/2)k^3x^3] $
Thus
$\displaystyle \lim_{x \to 0} \frac{sin [ sin ( sin[kx] ) ]}{x} \to \lim_{x \to 0} \frac{kx - (1/2)k^3x^3}{x}$ $\displaystyle \to k $
-Dan
In L'Hopitals rule you differentiate the top and bottom separately, andOriginally Posted by topsquark
the limit of the indeterminate form is the limit of the ratio of the derivatives.
As far as I can see the OP has found the ratio of derivatives correctly.
If you put the missing brackets in you will see that what she has does
in fact go to k as x goes to zero, as expected.
RonL